Question

设无穷等差数列An的前n项和为Sn,若首项a1=3/2,公差d=1,求满足S（k的平方）=(Sk)的平方的正整数k

Answer 1

sk=a1k+k(k-1)d/2=3k/2+k(k-1)/2=k(k+2)/2

sk^2=k^2(k^2+2)/2=(sk)^2=k^2(k+2)^2/4

2(k^2+2)=k^2+4k+4

k^2=4k

k=4

sk=a1k+k(k-1)d/2=k[a1+d(k-1)/2]

sk^2=k^2[a1+d(k^2-1)/2]=(sk)^2=k^2[a1+d(k-1)/2]^2

a1+d(k^2-1)/2=a1^2+a1d(k-1)+d^2(k-1)^2/4

k=1

a1=a1^2,a1=0 or 1

k=2,3d/2=a1d+d^2/4,d=0 or 6-4a1

a1=0,d=0 or 6

a1=1,d=0 or 2

d=0,为常数序列,a1=0,or 1 都满足.

d=2,a1=1,Sk=k^2,也满足

d=6,a1=0,an=6(n-1),sn=3(n-1)n,sn^2=3n^2(n-1)^2,s(n^2)=3n^2(n^2-1),两者不等.

因此只有上面三种情况